3.207 \(\int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=156 \[ -\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {4 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[Out]

(-4-4*I)*a^(5/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+4*a^2*(a+I*a*tan(d*x+c))^(
1/2)/d/tan(d*x+c)^(1/2)-2/3*I*a*(a+I*a*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(3/2)-2/5*(a+I*a*tan(d*x+c))^(5/2)/d/tan
(d*x+c)^(5/2)

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Rubi [A]  time = 0.28, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3548, 3545, 3544, 205} \[ \frac {4 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(7/2),x]

[Out]

((-4 - 4*I)*a^(5/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + (4*a^2*Sqrt[
a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (((2*I)/3)*a*(a + I*a*Tan[c + d*x])^(3/2))/(d*Tan[c + d*x]^(3/
2)) - (2*(a + I*a*Tan[c + d*x])^(5/2))/(5*d*Tan[c + d*x]^(5/2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3545

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m - 1)*(a*c - b*d)), x] + Dist[(2*a^2)/(
a*c - b*d), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f},
x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2]

Rule 3548

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f*m*(c^2 + d^2)), x] + Dist[a/(a*c - b*d), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*
d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n + 1, 0] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+i \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-(2 a) \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {4 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\left (4 i a^2\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {4 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {\left (8 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {4 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 3.56, size = 187, normalized size = 1.20 \[ -\frac {4 i a^2 e^{-i (c+d x)} \cot (c+d x) \left (e^{i (c+d x)} \left (-35 e^{2 i (c+d x)}+26 e^{4 i (c+d x)}+15\right )-15 \left (-1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (-1+e^{4 i (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(7/2),x]

[Out]

(((-4*I)/15)*a^2*(E^(I*(c + d*x))*(15 - 35*E^((2*I)*(c + d*x)) + 26*E^((4*I)*(c + d*x))) - 15*(-1 + E^((2*I)*(
c + d*x)))^(5/2)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*
x]])/(d*E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(-1 + E^((4*I)*(c +
d*x))))

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fricas [B]  time = 0.56, size = 465, normalized size = 2.98 \[ \frac {\sqrt {2} {\left (208 i \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} - 72 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 160 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 120 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 15 \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + i \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) - 15 \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - i \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right )}{30 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/30*(sqrt(2)*(208*I*a^2*e^(7*I*d*x + 7*I*c) - 72*I*a^2*e^(5*I*d*x + 5*I*c) - 160*I*a^2*e^(3*I*d*x + 3*I*c) +
120*I*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2
*I*c) + 1)) + 15*sqrt(32*I*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c)
- d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x
+ 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + I*sqrt(32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2) - 15
*sqrt(32*I*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*log(1/4*(4
*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(
e^(2*I*d*x + 2*I*c) + 1)) - I*sqrt(32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2))/(d*e^(6*I*d*x + 6*I
*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)/tan(d*x + c)^(7/2), x)

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maple [B]  time = 0.22, size = 414, normalized size = 2.65 \[ \frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (15 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{3}\left (d x +c \right )\right ) a +15 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{3}\left (d x +c \right )\right ) a +60 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, \left (\tan ^{3}\left (d x +c \right )\right ) a -22 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+76 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )-6 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{15 d \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x)

[Out]

1/15/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(15*I*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+
I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a+15*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)
*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a+60*ln(1
/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+
c)^3*a-22*I*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+76*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2-6*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I
*a)^(1/2))/tan(d*x+c)^(5/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)

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maxima [B]  time = 0.88, size = 1278, normalized size = 8.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

-1/225*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((-(900*I - 900)*a^2*cos(3*d*x
+ 3*c) + (930*I - 930)*a^2*cos(d*x + c) + (900*I + 900)*a^2*sin(3*d*x + 3*c) - (930*I + 930)*a^2*sin(d*x + c))
*cos(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + ((900*I + 900)*a^2*cos(3*d*x + 3*c) - (930*I + 93
0)*a^2*cos(d*x + c) + (900*I - 900)*a^2*sin(3*d*x + 3*c) - (930*I - 930)*a^2*sin(d*x + c))*sin(3/2*arctan2(sin
(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a) + (((900*I - 900)*a^2*cos(2*d*x + 2*c)^2 + (900*I - 900)*a^2*s
in(2*d*x + 2*c)^2 - (1800*I - 1800)*a^2*cos(2*d*x + 2*c) + (900*I - 900)*a^2)*arctan2((cos(2*d*x + 2*c)^2 + si
n(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - c
os(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*
x + 2*c), -cos(2*d*x + 2*c) + 1)) - sin(d*x + c)) + (-(450*I + 450)*a^2*cos(2*d*x + 2*c)^2 - (450*I + 450)*a^2
*sin(2*d*x + 2*c)^2 + (900*I + 900)*a^2*cos(2*d*x + 2*c) - (450*I + 450)*a^2)*log(cos(d*x + c)^2 + sin(d*x + c
)^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c)
, -cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2) - 2*(cos(2*d*x + 2*
c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c)
+ 1))*sin(d*x + c) + cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))))*(cos(2*d*x + 2
*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + (((900*I - 900)*a^2*cos(5*d*x + 5*c) - (7
50*I - 750)*a^2*cos(3*d*x + 3*c) + (210*I - 210)*a^2*cos(d*x + c) - (900*I + 900)*a^2*sin(5*d*x + 5*c) + (750*
I + 750)*a^2*sin(3*d*x + 3*c) - (210*I + 210)*a^2*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x +
2*c) + 1)) + ((-(240*I - 240)*a^2*cos(d*x + c) + (240*I + 240)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (240*I
- 240)*a^2*cos(d*x + c) + (-(240*I - 240)*a^2*cos(d*x + c) + (240*I + 240)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^
2 + (240*I + 240)*a^2*sin(d*x + c) + ((480*I - 480)*a^2*cos(d*x + c) - (480*I + 480)*a^2*sin(d*x + c))*cos(2*d
*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(900*I + 900)*a^2*cos(5*d*x + 5*c) +
(750*I + 750)*a^2*cos(3*d*x + 3*c) - (210*I + 210)*a^2*cos(d*x + c) - (900*I - 900)*a^2*sin(5*d*x + 5*c) + (75
0*I - 750)*a^2*sin(3*d*x + 3*c) - (210*I - 210)*a^2*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x
+ 2*c) + 1)) + (((240*I + 240)*a^2*cos(d*x + c) + (240*I - 240)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (240*I
+ 240)*a^2*cos(d*x + c) + ((240*I + 240)*a^2*cos(d*x + c) + (240*I - 240)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^
2 + (240*I - 240)*a^2*sin(d*x + c) + (-(480*I + 480)*a^2*cos(d*x + c) - (480*I - 480)*a^2*sin(d*x + c))*cos(2*
d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a))/((cos(2*d*x + 2*c)^2 + sin(2*d
*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(5/4)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(5/2)/tan(c + d*x)^(7/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^(5/2)/tan(c + d*x)^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(5/2)/tan(d*x+c)**(7/2),x)

[Out]

Timed out

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